Question

If $f(x)$ is a polynomial function satisfying $f(x)\cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$ and $f(4)=257$, then $f(3)=$

• A. 28
• B. 65
• C. 82
• D. 244

Answer 1

Answer: (C)

Let
$f(x)=a_0{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+\dots +a_{n-1}x+a_n$
Then, $f(x)\cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$
$\Rightarrow \left(a_0{x}^n+a_1{x}^{n-1}+\dots +a_n\right)$
$\left(\frac{a_0}{x^n}+\frac{a_1}{x^{n-1}}+\dots +a_n\right)$
On comparing the coefficient of $x^n$, we have
$a_0{a}_n=a_0$
$\Rightarrow a_n=1$
Comparing the coefficient of $x^{n-1}$, we have
$\Rightarrow a_0{a}_{n-1}+a_n{a}_1=a_1$
$\Rightarrow a_0{a}_{n-1}+a_1=a_1$
[ as $a_n=1]$
$\Rightarrow a_0{a}_{n-1}=0$
$\Rightarrow a_{n-1}=0$
[ as $a_0\ne 0]$
Similarly, $a_{n-1}=a_{n-2}=\dots =a_1=0$
and $a_0=\pm 1$
$\therefore f(x)=1\pm x^n$
$f(4)=1\pm 4^n=257$
$\Rightarrow 4^n=256$
$\Rightarrow 4^n=256$
$\Rightarrow 4=4$
$f(x)=1+x^4$
So, $f(3)=1+3^4=82$